How do I show that the series given by
$\sum_{n=1}^{\infty} \log\left(1+\frac{x}{n^3}\right)$ converges pointwise for $-1<x<1$.
I can't find a convergence test that works.
I have tried looking at:
$\sum_{n=1}^{\infty} { \left\lvert \log\left(1+\frac{x}{n^3}\right) \right\rvert} $
for $x=0$ the series is trivially convergent.
For $0<x<1$ I found you could use the series of $1/n^3$ where
$$\log\left(1+\frac{x}{n^3}\right) \leq \frac{x}{n^3} \leq \frac{1}{n^3}, $$
and then from the comparison test then our series will be absolute convergent and thereby convergent for positive $x$. But for negative $x$ i cannot find a series or test to compare it with. Any hints or proofs?

2 Answers
2

For $x \in (-1,1)$ you have$$\log \left(1+\frac{x}{n^3} \right) \leq \frac{x}{n^3}.$$
Consider the function $f(x) = \log(1+x)-x$. You have $f(0)=0$ and $f'(x) = -\frac{x}{1+x}$. So for $x>0$ your derivative $f'(x)$ is negative. For $-1<x<0$ you have $1+x > 0$, and so $f'(x)=-\frac{x}{1+x} > 0$. This means the function $f$ has a maximum at $x=0$. So $f(x) \leq 0$ for any $x\in (-1,1)$ in particular, that is
$$\log (1+x) \leq x, \quad x \in (-1,1).$$ This prevents our series from diverging to $+\infty$.

Further, observe that for $x \in (-1,0)$
$$\lim_{n \rightarrow +\infty} \frac{\log \left(1+\frac{x}{n^3} \right)}{\frac{x}{n^3}} = 1.$$ Thus if you fix $\varepsilon > 0$ there is a natural number $N$ such that $$(1-\varepsilon)\frac{x}{n^3} < \log \left(1+\frac{x}{n^3} \right) < (1+\varepsilon)\frac{x}{n^3}$$
for any $n \ge N$.
Now fix $\varepsilon > 0$ such that $1-\varepsilon > 0$. Then there is $N$ such that
$$-\infty < (1-\varepsilon)x\sum_{n=N}^{+\infty}\frac{1}{n^3} < \sum_{n=N}^{+\infty}\log \left(1+\frac{x}{n^3} \right) < +\infty.$$
All of this prevents our series from diverging to $-\infty$ as well.